Question

A 60 Hz generator, connected directly to an infinite bus operating at a voltage of $1\angle 0^{\circ }$ p.u., has a synchronous reactance of 1.35 p.u. The generator no load voltage is 1.1 p.u. and its inertia constant H is 4 MJ/MVA. The generator is suddenly loaded to 60 percent of its maximum power limit. The frequency of the resulting natural oscillations of the generator rotor is

• A. 1.246 Hz
• B. 0.882 Hz
• C. 0.921 Hz
• D. 1.16 Hz

Answer 1

The correct option is B 0.882 Hz

$0.6P_{max}=P_{max}\sin \delta$
$delta={\sin }^{-1}\left(0.6\right)=36,{8698}^{\circ }$
${\begin{array}{c}\frac{\partial P_e}{\partial \delta }\end{array}|}_{\delta ={36.8699}^{\circ }}=P_{max}\cos \left({36.8698}^{\circ }\right)$
$=\frac{1.1\times 1}{1.36}\cos \left({36.8698}^{\circ }\right)=0.6518p.i./rad$
Frequency of oscillations $=\sqrt{\frac{{\left(\frac{\partial P_e}{\partial \delta }\right)}_{{36.8699}^{\circ }}}{M}}$
Where, $M=\frac{H}{\pi f}=\frac{4}{\pi \times 60}=0.0212p.u.s^2/rad$
$f=\sqrt{\frac{0.6518}{0.0212}}$
$f={5.54}^{\prime }\text{rad/sec}=0.882Hz$